(5+1/6+)x+(2+1/4)=10

Solution for (5+1/6+)x+(2+1/4)=10 equation:

(5+1/6+)x+(2+1/4)=10

We move all terms to the left:

(5+1/6+)x+(2+1/4)-(10)=0


Domain of the equation: 6+)x!=0

We move all terms containing x to the left, all other terms to the right

)x!=-6

x!=-6/1

x!=-6

x∈R


determiningTheFunctionDomain
(5+1/6+)x-10+(2+1/4)=0

We add all the numbers together, and all the variables

(+1/6)x-10+(1/4+2)=0

We multiply parentheses

x^2-10+(1/4+2)=0

We get rid of parentheses

x^2-10+2+1/4=0

We multiply all the terms by the denominator


x^2*4+1-10*4+2*4=0

We add all the numbers together, and all the variables

x^2*4-31=0

Wy multiply elements

4x^2-31=0

a = 4; b = 0; c = -31;
Δ = b2-4ac
Δ = 02-4·4·(-31)
Δ = 496
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{496}=\sqrt{16*31}=\sqrt{16}*\sqrt{31}=4\sqrt{31}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{31}}{2*4}=\frac{0-4\sqrt{31}}{8}
=-\frac{4\sqrt{31}}{8}
=-\frac{\sqrt{31}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{31}}{2*4}=\frac{0+4\sqrt{31}}{8}
=\frac{4\sqrt{31}}{8}
=\frac{\sqrt{31}}{2}
$